Find the volume between the sphere of radius 1 and the cylinder of radius $1/2$ above the plane $xy$. Both are centered at the origin.
I'm not so sure on how to apply spherical coordinates to solve this problem. Calculating the volume of the cone is pretty easy; it's parametrization is given by $$\theta\in[0,2\pi], r\in[0,1], \varphi\in[0,\pi/6],$$ where $\theta$ is the azimuth and $\varphi$ the inclination. To obtain the value of $\varphi$ it's straight forward since $x^2+y^2=1/4$, $x^2+y^2+z^2=1$ and $z = r\cos\varphi$ considering $r=1$.
Now, I'm stuck on how to find the other possible values of $\theta, \varphi$ and $r$ that represent the complement of the cone intersection the cylinder. Well, mostly I don't know how to find $r$, since it's clear that $\theta\in[0,2\pi]$ and that $\varphi\in[\pi/6,\pi/2]$ by the above conclusion on the cone and how it works spherical coordinates. Any ideas on how to find the possible values of $r$?
The complement of the intersection is $~0\leqslant\theta\leqslant 2\pi~~,~~0\leqslant \varphi\leqslant \pi/6~$, and $~1/2\leqslant r\sin\varphi \leqslant\sin\varphi~.$
Oh, and the volume of the whole hemisphere is $2\pi/3$ .
Sketch the x-z cross section of the hemisphere and a diagonal from the origin at some angle ($\varphi$) crossing the surface of the hemisphere between the x-axis and the intersection with the cylinder (ie $\varphi\in[0:\pi/6]$. See how the length along the diagonal from the origin to the cylinder is $1/(2\sin(\varphi)$ . [ie: the length of the hypotenuse for a right triangle with base $1/2$ and adjacent angle $\varphi$.]